Joint distribution of X and Y bernoulli random variables

Question

A box contains two coins: a regular coin and a biased coin with $P(H)=\frac23$. I choose a coin at random and toss it once. I define the random variable X as a Bernoulli random variable associated with this coin toss, i.e., X=1 if the result of the coin toss is heads and X=0 otherwise. Then I take the remaining coin in the box and toss it once. I define the random variable Y as a Bernoulli random variable associated with the second coin toss. a)Find the joint PMF of X and Y.

b)Are X and Y independent?

My attempt to answer this question:

Let A be the event that first coin, I pick is the regular(fair) coin. Then conditioning on that event, I can find joint PMF. Once conditioned, I can decide if X and Y are independent(conditionally).

$P(A)=\frac12,P(A^c)=\frac12$.

In the event A, $P(X=1)=\frac12,P(Y=1)=\frac23$.

In the event$A^c, P(X=1)=\frac23, P(Y=1)=\frac12$

So, $P_{X,Y}(x,y)= P(X=x, Y=y|A)P(A) + P(X=x, Y=y|A^c)P(A^c)$

$P_{X,Y}(x,y)=P_{\frac12}(x)P_{\frac23}(y)(\frac12) + P_{\frac23}(x)P_{\frac12}(y)(\frac12)$

Now, how can we find Joint PMF of X and Y using Bernoulli distribution?

Answer 1

The joint pmf can be described by a 2-by-2 contingency table that shows the probabilities of getting $X=1$ and $Y=1$, $X=1$ and $Y=0$, $X=0$ and $Y=1$, $X=0$ and $Y=0$. So you'll have:

	$X=0$	$X=1$
$Y=0$	$\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{2}=\frac{1}{6}$	$\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{1}{2}=\frac{1}{4}$
$Y=1$	$\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{2}{3}+\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{2}=\frac{1}{4}$	$\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{2}{3}+\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{1}{2}=\frac{1}{3}$

Answer 2

Let $U$ take value $1$ if the regular coin results in heads and let $U$ take value $0$ otherwise.

Let $V$ take value $1$ if the biased coin results in heads and let $V$ take value $0$ otherwise.

Then by independence of $U$ and $V$:$$P(X=1,Y=1)=P(U=1,V=1)=P(U=1)P(V=1)=\frac12\frac23=\frac13$$ and: $$P(X=0,Y=0)=P(U=0,V=0)=P(U=0)P(V=0)=\frac12\frac13=\frac16$$ Further by symmetry:$$P(X=1,Y=0)=P(X=0,Y=1)$$ so that we can conclude that:$$P(X=1,Y=0)=P(X=0,Y=1)=\frac12\left(1-\frac13-\frac16\right)=\frac14$$